# Tag Info

0

Suppose that $(f,f_{n})=0$ for all $n$. Then $f=0$ because, if not, $$\|f\|^{2}=\sum_{n}|(f,g_{n})|^{2}=\sum_{n}|(f,g_{n}-f_{n})|^{2} \le \sum_{n}\|f\|^{2}\|f_{n}-g_{n}\|^{2}< \|f\|^{2}.$$ Therefore the orthonormal set $\{ f_{n}\}$ must be complete because there is no non-zero vector $f$ which is orthogonal to every $f_{n}$.

1

Let $F$ be the scalar field. Since $f\ne 0$, there is a $v\in H$ such that $f(v)\ne 0$; let $\alpha=f(v)$. For surjectivity let $\beta\in F$ and consider $$f\left(\frac{\beta}{\alpha}v\right)\;.$$ For the rest, note that if $\dim H>1$, then $\ker f\ne(0)$ (why?), so pick any non-zero $u\in\ker f$ and consider $f(u+v)$.

0

Careful investigations show: $$\|f(E)\|<\infty\implies f\in\mathcal{B}(E)\implies f(E)\in\mathcal{B}(\mathcal{H})$$ (I will post a proof of this as soon I will find some time!) That gives the equivalence: $$r(E)<\infty\iff\mathrm{id}\in\mathcal{B}(E)\iff T\in\mathcal{B}(\mathcal{H})\iff\|T\|<\infty$$ Moreover, it holds: $$\sigma(T)=\mathrm{supp}E$$ ...

0

For the functional calculus it holds rigorously: $$\mathcal{N}(f(E))=\{0\}\iff E(\{f=0\})=0$$ $$f(E)\in\mathcal{B}(\mathcal{H})\iff f\in\mathcal{B}(E)$$ and especially: $$f(E)^{-1}=f^{-1}(E)$$ $$\|f(E)\|=\|f\|$$ (I will post a proof for this as soon I find some time!) Exploiting this one has: \begin{align}\lambda\in\rho(T)&\iff ...

0

I just realized that this cannot be true for the following reason: Assume every dense subspace would provide an ONB for the Hilbert space: $$\overline{\mathcal{S}}=\mathcal{H}\quad(\mathcal{S}\subseteq Z)$$ Then, it would serve as well as an ONB for the subspace: $$\overline{\mathcal{S}}\supseteq Z$$ But there are preHilbert spaces which do not admit any ...

1

Let $z\in X$. we define $f_{z}(x)=<x,z>$. f is linear: $(f_{z}(ax+by)=<ax+by,z>= a<x,z>+b<y,z>=af_{z}(x)+bf_{z}(y)$ f is bounded: $|f_{z}(x)|=|<x,z>|\leq\|x\|\|z\|\Longrightarrow \|f\|\leq\|z\|$ (where the first inequality is the Cauchy-swartz inequality). Let $T: X\longrightarrow X^*$ with $T(z)=f_{z}$. Observe that $T$ is ...

-1

An example is provided by nondiagonalizable matrices: (Thanks @user161825!!) Consider the matrix: $$N:=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ and the subspace: $$Z:=\{\begin{pmatrix}a\\0\end{pmatrix}:a\in\mathbb{C}\}$$ Then it fails to reduce: $$NZ\subseteq Z$$ $$NZ^\perp\nsubseteq Z^\perp$$

1

Remember that $E(A)^{2}=E(A)=E(A)^{\star}$. So $$\mu_{x,x}=(E(A)x,x)=(E(A)^{2}x,x)=(E(A)x,E(A)^{\star}x)=(E(A),E(A)x)=\nu_{x}.$$ It is interesting that $\mu_{x,y} << \mu_{x,x}$ and $\mu_{x,y} << \mu_{y,y}$ because $$|\mu_{x,y}(A)| \le \mu_{x,x}(A)^{1/2}\mu_{y,y}(A)^{1/2}.$$ Radon-Nikodym gives you something interesting, ...

0

As Paul remarked above- this is not true. Take $h_1$ and $h_2$ to be two orthogonal vectors in $\mathcal{H}$, and $T_1=T_2=I$. Then, your condition 1) holds, but condition 2) doesn't.

1

For any orthogonal projection $P$, a vector $x$ is in the range of $P$ iff $\|Px\|=\|x\|$. The condition $P \le Q$ for orthogonal projections is equivalent to $\|Px\|\le \|Qx\|$. Because $\|Qx\|\le \|x\|$ it follows that if $x \in\mathcal{R}(P)$, then $x\in\mathcal{R}(Q)$.

1

Suppose $E$ is a spectral measure on $\mathbb{C}$. Choose any unit vector $x \in \mathcal{H}$, and look at the probability measure $\mu_{x}(S)=\|E(S)x\|^{2}$. If there are countably many disjoint Borel subsets $\{ S_{j}\}_{j=1}^{\infty}$ with $\mu_{x}(S_{j})\ne 0$, then you can construct a Borel function $f$ such that $\int |f|^{2}d\mu_{x} =\infty$. Then ...

1

Not true. Take $f(x,y)=x+iy$ and $X$ the space spanned by $f(x)$ and $\overline{f}(x,y)=x-iy$ over $\mathbb C$. Now, every element o $X$ is of the form $u=af+b\overline{f}$. Define the inner product $$\langle u_1,u_2\rangle=a_1\overline a_2+4 b_1\overline b_2.$$ Then $$\|\,\overline f\|=2\|\,f\|.$$

2

I believe I have a counterexample to your statement about the equality $\|x\|=\|\bar{x}\|$. Let $S=\{a, b\}$ be a set with two entries. Define a complex-valued function $f$ on $S$ by $$f(a)=1+i, \quad f(b)=1$$ and consider the two-dimensional space $V=\operatorname{Span}\{f, \bar{f}\}$ over $\mathbb{C}$ equipped with the inner product $$(\phi, ... 0 I finally got this to work, so I'd like anyone looking for answeres here to know I made a mistake in (1); it should say: \left(\frac{d^n}{dt^n}(t^2-1)^n\right) = \sum _{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)\left(\prod _{j=0}^{n-1}(2n-2k-j)\right)t^{2n-2k}(-1)^k = \sum ... 2 It is often difficult to prove that an operator is self-adjoint, which seems reasonable since you know a lot about an operator when it is self-adjoint. For instance it is unitarily equivalent to a multiplication operator by a real function. There are many possible approaches, one which is sometimes useful is the following Show that D is symmetric, i.e. ... 0 Let H = {\left( {{L^2}\left( {{\mathbb{R}^3},{\mathbb{C}^4}} \right)} \right)^4} be the Hilbert space in question and for u = \left( {{u_1},{u_2},{u_3},{u_4}} \right),v = \left( {{v_1},{v_2},{v_3},{v_4}} \right) \in H let the inner product on H be defined by \left\langle {u,v} \right\rangle = \sum\limits_{i = 1}^4 {\int_{ - \infty }^{ + \infty } ... 0 Let \langle \cdot, \cdot \rangle be the dot product and T be the operator taking e_1 to e_2 and e_2 to -e_1, i.e.$$T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$So we have$$\left\langle T\begin{pmatrix}\alpha \\ \beta \end{pmatrix}, \begin{pmatrix}\alpha \\ \beta \end{pmatrix}\right\rangle = \left\langle \begin{pmatrix}-\beta \\ ...

1

I now looked up the definition of a core of an operator and it could be that the answer actually depends on the exact definition. In Wikipedia, a core (of a closed operator) is defined as a subset $D$ of the domain of $A$ such that $A$ is the closure of $A|D$. If we replace this by requiring that $D$ be a subspace, I can prove your claim. By symmetry, it ...

1

Suppose that $a\ne0$, and consider first the case of continuous functions $f\in L^2([0,1])$, in this case $(e^{at}\phi(t))'= e^{at}f(t)$. So, since $\phi(0)=0$, $$e^a\phi(1)=\int_0^1e^{at}f(t)dt\tag{1}$$ Now, integrating $\phi'+a\phi=f$ on $[0,1]$, we get $$\phi(1)+a\,l(f)=\int_0^1f(t)dt\tag{2}$$ Combining $(1)$ and $(2)$ we obtain $$... 1 The isomorphism H \to H^* is always there, whether you pay attention to it or not. Making it explicit would be annoying; it would be akin to disallowing yourself to treat a rational number as if it were a real number, and having to add extra notation whenever you want to convert a rational number to its corresponding real number. 1 In general, it does not work. Construct \mathcal F=\{z_n, \ n\in\mathbb N\} as the set of sequences defined in the following way: z_n=(z_{n,k}) will be the sequence of elements in l^2 with$$ z_{n,k} = \begin{cases}\frac1n e_{1} + \frac{\sqrt{n^2-1}}ne_{2} & \text{ if } k=1\\ \frac1n e_{2} + \frac{\sqrt{n^2-1}}ne_{k+1} & \text{ if } ...

1

The commutation relation $[A,B]=iI$ (i.e., $AB-BA=iI$) is not satisfied by two linear operators on any finite-dimensional linear space. So Quantum Mechanics requires infinite-dimensional spaces. Even in infinite-dimensional spaces, the commutation relation cannot be satisfied by two bounded operators $A$ and $B$; at least one of them must be discontinuous. ...

0

From Lagrange multiplies the following Lagrange equation can be immediately derived: $$g(x)=a x^3+bx^2+cx+d$$ This equation is not differential because the functional does not depend on $g^\prime$. The constant $a,b,c,d$ are derived from the constraints on $g$. Indeed we have: $$2b/3+2d=0\\ a/2+c=0\\ 2b/5+2d/3=0\\$$ Then $b=d=0$ and $a=-2c$ so ...

0

Let $H$ be a Hilbert space and let $M$ be a closed subspace of $H$. If $x \in X$, then $$\sup_{z \in M^\perp, \|z\| = 1} |\langle x,z \rangle| = \inf_{y \in M} \|x-y\|$$ because if $y \in M$ and $z \in M^\perp$ with $\|z\| = 1$ you have $$|\langle x,z \rangle| = |\langle x-y,z \rangle| \le \|x-y\|$$ since $\langle y,z \rangle = 0$. On the other hand, you ...

1

(1) By the definition of infimum, $d+\frac1n$ is not a lower bound for $\{\|x-y\|^2 : y\in M\}$, that is, there exists $y_n\in M$ such that $\|x-y_n\|^2<d+\frac1n$. Then $d\le \|x-y_n\|^2 < d+\frac1n$; apply the squeeze theorem. (2) $M$ is a subspace, so it's closed under linear combinations.

2

I am not a fan of the term 'hermitian', because it is so often used haphazardly (sometimes it means symmetric, sometimes it means self-adjoint, sometimes it presupposes that the vector space is finite-dimensional). The term 'self-adjoint' seems to be used much more consistently, to mean a densely defined operator which satisfies $A=A^*$ (equivalently $A$ is ...

1

In the finite case every operator is bounded and since the domain is the whole Hilbert space, every hermitian operator is self-adjoint. For the infinite case, the good definition of a hermitian operator is a densely defined operator $T$ such that $T\subset T^*$, i.e., it is symmetric and the adjoint $T^*$ is an extension, we have ...

1

I'm going to assume the facts that I proved in a previous answer to a question of yours: Spectral Measure Integration: Unbounded Functions? For any Borel function $f$, I defined an operator $T_{f}$ on $\mathcal{D}(T_{f})$ consisting of all $x \in \mathcal{H}$ for which $\int |f(\lambda)|^{2}\,d\|E(\lambda)x\|^{2}<\infty$. I showed that $T_{f}$ is a ...

0

Suppose that $t_0\in \operatorname{supp}E$, that is $E(B(t_0,\varepsilon))\neq0, \forall\varepsilon>0$. Then, there exists a normalized sequence $(x_j)$, with $$x_j\in E(B(t_0,1/j))\mathcal{H}, \qquad \forall j\in\mathbb{N}.$$ Hence $E(B(t_0,1/j))x_j=x_j$, and it follows that $\mu_{x_j}^T\left(\Bbb{C}\backslash B(t_0,1/j)\right)=0$. Thus ...

0

The equality holds, including the infinity case. For normal bounded operators, since $\|T\|=\|T^*\|$ we verify that $$\|T\|^2=\sup_{\|\xi\|=1}\|T\xi\|^2=\sup_{\|\xi\|=1}\langle T\xi,T\xi\rangle =\sup_{\|\xi\|=1}\langle T^*T\xi,\xi\rangle =\|T^*T\|\leq\|T\|\|T^*\|=\|T\|^2,$$ and, $\|T^*T\|=\|T\|^2$. But we also have $$\|T^*T\xi\|^2=\langle ... 0 here is my proof. (I used the biadjoint) My proof 1 This is impossible. For a proper convex and lower semicontinuous function the domain of definition of the subdifferential is dense in the domain of f,$$ \mathrm{dom}(f) = \overline{ D(\partial f)}. $$This implies that there is x\in X such that \partial f(x)\ne\emptyset, as for proper convex and lower semicontinuous f it holds \mathrm{dom}(f) \ne ... 0 Hint: there is a subsequence all of whose coordinates converge. Show that the sequence converges to the vector whose coordinates are these limits. Details: 2 A Hilbert space has been identified with its dual, if at places where one would expect an element from H^* instead an element of H is written. Let me explain this with some examples. (1) Consider a function f:H\to \mathbb R. If f is Gateaux differentiable, then its first derivative f'(x) is an element in H^*: it maps a direction (in H) to an ... 1 Note that$$|\langle Ax, y\rangle |\le \|Ax\|\cdot \|y\|.$$So,$$\frac{|\langle Ax, y\rangle |}{\|x\|\cdot \|y\|}\le \frac{\|Ax\|}{\|x\|}.$$Thus$$||L|| = \sup_{x,y \in H, x,y \neq 0} \frac{|\langle Ax, y\rangle|}{||x||\cdot ||y||}\le \sup_{x\in H, x \neq 0}\frac{\|Ax\|}{\|x\|}. $$On the other hand, for x=y you get the desired equality. 0 Two things to be checked: a. They are orthonormal, indeed$$ \int_{-\pi}^{\pi}\mathrm{e}^{kx i}\overline{\mathrm{e}^{\ell x i}}\,dx= \int_{-\pi}^{\pi}\mathrm{e}^{kx i}{\mathrm{e}^{-\ell x i}}\,dx= \int_{-\pi}^{\pi}\mathrm{e}^{(k-\ell)x i}=\left\{\begin{array}{ccc} 2\pi & \text{if} & k=\ell,\\ 0 & \text{if} & k\ne\ell. \end{array}\right. $$... 2 If you know about compact selfadjoint operators, then you can study the operator$$ Lf = \frac{1}{i}\frac{d}{dt} $$on the domain \mathcal{D}(L) consisting of all absolutely continuous functions f on [0,2\pi] with f(0)=f(2\pi) and f' \in L^{2}[0,2\pi]. L is not compact, but its resolvent operator (\lambda I-L)^{-1} ... 1 Stone-Weierstrass, basically. You can check that these functions form an algebra that separates points, contains constant functions, and is closed under complex conjugation. The interval is compact, thus this algebra is dense in C([-\pi,\pi]) in the supremum norm, thus in the L^2 norm. Further, the continuous functions are dense in L^2, so this forms ... 1 First note that if \langle Ah, h'\rangle=0 for all h,h'\in\mathbb H then A=0. Recall the polarization identity$$\langle h,h'\rangle=\frac{1}{4}(\langle h+h',h+h'\rangle - \langle h-h',h-h'\rangle +\langle h+ih',h+ih'\rangle-\langle h-ih',h-ih'\rangle$$It follows from the first observation and the polarization identity that \langle Ah,h\rangle=0 for ... 2 This is not true:$$f\in L^1(\kappa)\iff f\in L^1(\lambda).$$However, from \kappa(E)\le c\lambda(E), you can deduce that$$ f\in L^1(\lambda)\quad\Longrightarrow\quad f\in L^1(\kappa). $$2 As Daniel Fischer noted, condition (P) is equivalent to the norm being strictly convex. Indeed, if the norm is not strictly convex, then the unit sphere contains a line segment, and removing the midpoint of this line segment from the closed unit ball creates a nonconvex set. Conversely, if the norm is strictly convex, then any X as you described is ... 0 If A : \mathcal{D}(A)\subseteq H\rightarrow H is symmetric on its domain, then A is selfadjoint iff A\pm iI are surjective. If these operators are surjective, then the domain is automatically dense, which saves some checking. 1 The form is <\phi,\phi>_{A}=\|A^{1/2}\phi\|^{2}_{L^{2}}+\|\phi\|^{2}_{L^{2}} is defined on \mathcal{D}(A), a subspace consisting of all \phi \in L^{2} such that A\phi \in L^{2}. Notice that the form norm is the same as the graph norm of A^{1/2}. So you want to show that the restriction A^{1/2}|_{\mathcal{D}(A)} of A^{1/2} has closure ... 1 Suppose H has a countably dense subset \{ s_{n}\}_{n=1}^{\infty} of non-zero vectors. Using induction we may discard each s_{n} which is a linear combination of the previous vectors in the sequence in order to arrive at a new sequence \{ t_{n}\}_{n=1}^{\infty} of vectors for which the \{ t_{1},t_{2},\cdots,t_{k}\} is linearly independent for each ... 2 I take it by "complete" you mean that the f_n have dense linear span or, equivalently, they have zero orthogonal complement. This we can see as follows. Suppose that \langle g, f_n\rangle =0 for all n. Then$$ |\langle g, e_n\rangle | =|\langle g, e_n-f_n\rangle | \le \|g\|\, \|e_n-f_n\| . $$Take squares and sum over n. This yields \|g\|^2 \le S ... 2 In finite dimensional complex spaces, you can find a branch of complex log function that doesn't intersect any of the eigenvalues. And that offers one explanation of why everything works so nicely. You can see it by using Cauchy's integral formula$$ \log(A) = \frac{1}{2\pi i}\oint_{C} \frac{1}{\lambda I-A}\log(\lambda)\,d\lambda. $$You can prove that ... 0 Every Hilbert space has a natural notion of distance (metric), namely$$d(a,b)=\|a-b\| (This is also true for every normed space; indeed, one of the axioms of a norm is essentiall the triangle inequality for $d$.) Starting with the definition of Hilbert space norm as $\|x\|^2=\langle x,x\rangle$, one can prove the triangle inequality by using ...

1

I agree that, if you make the direct sum of a finite number of Hilbert spaces, then you do not need to complete anything. Indeed, the finite product of complete metric spaces is also complete. However, things change quite a little bit when you switch to infinite products: indeed, the algebraic direct sum of an infinite family of vector spaces consists not on ...

1

In your formulation the problem is not correctly posed. The operator $L$ is not defined for all $u\in \mathcal{H}=L^{2}(\mathbb{R}^{2},dxdy)$. But it is well-defined on a suitable dense set. Let $p_{x}$ and $p_{y}$ the self-adjoint momentum operators derived from $-i\partial _{x}$ and $% -i\partial _{y}$. Then on the domain of $xp_{x}p_{y}^{2}$ ...

0

Whether or not $z$ is real, if $R_{z}$ and $R_{\overline{z}}$ are assumed to be defined on all of $\mathcal{H}$, then these operators are closed because the graphs of $L-z I$, $L-\overline{z}I$ are closed and the graphs of $R_{z}$, $R_{\overline{z}}$ are their respective transposes in $\mathcal{H}\times\mathcal{H}$ and, hence, closed; so $R_{z}$, ...

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